Math 2260 Exam #1 Solutions
1. (12 points) The curves determined by the equations y = x2 and x = y 2 enclose a region in the first
quadrant, as seen below.
(a) What is the area of this region?
2
Answer: First, notice
√that the portion of the curve x = y adjacent to the region of interest also
lies on the curve
√ y = x. The points of intersection of the two curves will thus be determined by
where x2 = x, which is certainly where x = 0 and x = 1. Therefore, the points of intersection
are (0, 0) and (1, 1).
√
Since the curve y = x lies above y = x2 between x = 0 and x = 1, the area of this region is
equal to
Z
1
√
0
1
2 3/2 x3
x − x2 dx =
x −
3
3 0
2 1
= −
3 3
1
= .
3
(b) Rotating this region around the x-axis yields a solid. Write down the integrals which will determine
the volume of this solid using (i) the washer method and (ii) the shell method.
(i) Washer method integral:
Answer: Since we’re rotating around the x-axis, the washers will be vertical, so they change
as x changes. Therefore, we should express the area of a cross-sectional
washer as a function
√
of x. The outer radius of the washer is given by the curve y = x and the inner radius by
the curve y = x2 , so the area of the washer is
A(x) = π
2
√ 2
x − π x2 = πx − πx4 .
Therefore, the volume of the solid is
Z 1
Z
A(x) dx =
0
0
1
1
πx − πx4 dx.
(ii) Shell method integral:
Answer: Cylindrical shells, on the other hand, will be horizontal, changing as y changes, so
we should integrate with respect to y. Thus, we need to express the two curves as graphs of
√
functions of y: we already know that one is x = y 2 , and the other is obviously x = y. Since
√
the sides of a cylindrical shell run from the curve x = y to x = y 2 , the volume of the solid
will be given by the integral
Z 1
Z 1 √
2
2πy y − y dy =
2π y 3/2 − y 3 dy.
0
0
(c) Compute the volume of this solid by picking either of the integrals you wrote down in part (b)
and evaluating it.
Answer: Using the washer method integral from (b)(i), we see that the volume of the solid is
Z
1
4
πx − πx
1
Z
x − x4 dx
dx = π
0
0
x2
x5
−
2
5
1 1
=π
−
2 5
3π
=
.
10
1
=π
0
If instead I used the shell method integral from (b)(ii), I get a volume of
Z
1
Z
2π y 3/2 − y 3 dy = 2π
0
1
y 3/2 − y 3 dy
0
2 5/2 y 4
y
−
5
4
2 1
−
= 2π
5 4
3
= 2π
20
3π
=
.
10
1
= 2π
0
Either way, I get the same volume for the solid: 3π
10 .
√
2. (5 points) Between x = 0 and x = π, the curve y = sin x2 and the x-axis enclose a region. What
is the volume of the solid obtained by revolving this region around the y-axis?
2
Answer: Since we’re rotating around the y-axis, a washer would run horizontally across the region
whereas a cylindrical shell would have vertical sides. Based on the picture of the region shown above,
the cylindrical shell is clearly preferable.
Since the shells are vertical, they change as x changes, so we should express everything in terms of x.
Since we’re rotating
the radius of a shell based at x will be x − 0 = x, and it will
around the y-axis,
have height sin x2 − 0 = sin x2 . Therefore, the volume of the solid is equal to
Z
√
π
2πx sin x2 dx.
0
I will evaluate this integral using a u-substitution, with u = x2 . Then du = 2x dx and the above
integral is equal to
√
Z
π
π
2x sin x2 dx = π
Z
π
sin(u) du
h0
iπ
= π − cos(u)
0
0
= π [− cos(π) + cos(0)]
= 2π,
so the volume of the solid is 2π.
3. (5 points) What is the length of the segment of the curve y =
x3/2
2
between x = − 16
9 and x = 0?
Answer: Since I’m going to use the arc length integral, let me first determine the derivative of y:
1 3 1/2
3√
dy
=
x
=
x.
dx
22
4
Therefore, the length of the curve is
s
s
2
2
Z 0
Z 0
dy
3√
1+
dx =
1+
x dx
dx
4
− 16
− 16
9
9
Z 0 r
9
=
1 + x dx.
16
16
− 9
3
Now, let u = 1 +
9
16 x.
9
Then du = 16
dx and the above integral is equal to
Z 0 r
Z
9 9
16
16 1 √
1+ x
u du
dx =
9 − 16
16 16
9 0
9
1
16 2 3/2
=
x
9 3
0
16 2
−0
=
9 3
32
=
,
27
so the length of this segment of the curve is
32
27 .
2
4. (5√points)
Revolving the portion of the graph of the function f (x) = x between the points (0, 0) and
2, 2 around the y-axis yields a type of surface called a paraboloid. What is the surface area of this
paraboloid?
Answer: There are two ways to compute this surface area, both yielding the same answer.
First, notice that, since we’re revolving around the y-axis, the distance from a point on the curve
y = x2 to the axis of rotation is equal to x − 0 = x. Hence, the surface area will be given by
s
2
Z √2
Z √2
q
dy
2
2πx 1 +
dx =
2πx 1 + (2x) dx
dx
0
0
Z √2
p
=
2πx 1 + 4x2 dx.
0
Letting u = 1 + 4x2 means that du = 8x dx and the above integral is equal to
Z √2 p
Z
1
π 9√
2π
8x 1 + 4x2 dx =
u du
8 0
4 1
9
π 2 3/2
=
u
4 3
1
π 2
2
=
· 27 − · 1
4 3
3
π
2
=
18 −
4
3
π 52
=
4 3
52π
=
12
13π
=
.
3
On the other hand, if we express everything in terms of y, on the region we’re interested in the curve
√
√
is also given by x = y. Then the distance from the curve to the axis of rotation is equal to y and
so the surface area is equal to
s
s
2
2
Z 2
Z 2
dx
1
√
√
2π y 1 +
dy =
2π y 1 +
dy
√
dy
2 y
0
0
r
Z 2
1
√
= 2π
y 1+
dy.
4y
0
4
Combining the square roots yields
Z
2
r
2π
y+
0
Letting u = y +
1
4
1
dy.
4
means that du = dy and the above integral is equal to
Z
9/4
2π
1/4
√
9/4
2 3/2
u du = 2π u
3
1/4
2 27 2 1
= 2π
−
3 8
38
54
2
= 2π
−
24 24
52
= 2π
24
52π
=
12
13π
.
=
3
Either way, the surface area of the paraboloid is equal to
5
13π
3 .